1. ## Maximizing volume

I tried to work this out and ran into some problems with my answer, some guidance would be excellent.

You are planning to make an open rectangular box from a 12-cm by 14-cm piece of cardboard by cutting congruent squares from the corners and folding up the sides.

a) What are the dimensions of the box of largest volume you can make this way?

b) What is its volume?

This is my work:

V = (L)(W)(H)
L = 14 - x
W = 12 - x
H = x

V(x) = x(14-2x)(12-2x)
= x(4x^2-52x+168)
= X^3-13x^2+42x
I took the derivative and got
3x^2-26x+42 = 0
I solved with the quadratic formula and got x= 6.5 and x=2.15???

I am pretty sure i did something wrong, any help would greatly appreciated.
Thanks!

2. Originally Posted by Murphie
I This is my work:

V = (L)(W)(H)
L = 14 -$\displaystyle {\color{red}2}$ x
W = 12 -$\displaystyle {\color{red}2}$ x
H = x

V(x) = x(14-2x)(12-2x)
= x(4x^2-52x+168)
= X^3-13x^2+42x
I took the derivative and got
3x^2-26x+42 = 0
I solved with the quadratic formula and got x= 6.5 and x=2.15???

I am pretty sure i did something wrong, any help would greatly appreciated.
Thanks!
Those values for $\displaystyle x$ are not correct, have another go at the quadratic formula.

3. 3x^2 - 26x +42 + 0

(26 + √(676-504))/(6) = 33,
(26 - √(676-504))/(6) = 24.33
x=33
x=24.33
wow I can't believe I messed the last step up, do these look right then? and if so, does that mean that the Length 14-x is 2 numbers, 14-33 and 14-24.33?

4. You need to reject values outside $\displaystyle 0<x<6$

On reflection your first 2 solutions are correct $\displaystyle x \approx 2.15$